A Star Is Born! A Star Is Born! p. 1/3 A Star Is Born! The - - PowerPoint PPT Presentation

A Star Is Born!

A Star Is Born! – p. 1/3

A Star Is Born!

The photograph below shows a cloud of molecules called Bernard 68 (B68). It is located about 300 light-years (2.8 × 1015 km) away from us in the constellation Ophiuchus and is about 1.6 trillion kilometers across. It is made of molecules like CS, N2H, H2, and CO and is slowly rotating (ω = 9.4 × 10−14 rad/s). The internal gravitational attraction of B68 may make the molecular cloud collapse far enough so it will ignite the nuclear fires and B68 will begin to shine.

A Star Is Born! – p. 2/3

A Star Is Born

The molecular cloud B68 in the constellation Ophiuchus is rotating with an angular speed ω = 9.4 × 10−14 rad/s. The gravitational attraction among the atoms in the cloud may make it collapse until the core is hot enough to ignite nuclear reactions and B68 will begin to shine. If the final properties of B68 are the same as our Sun, i.e., the same mass and size, then what will be its final angular velocity and period? Assume the lost mass carries away very little angular momentum. Compare this with the angular velocity of the Sun. Is your result reasonable? Why or why not? MB68 = 6.04 × 1030 kg IB68 = 2.7 × 1054 kg − km2 MSun = 1.989 × 1030 kg RSun = 6.96 × 105 km TSun = 25.4 d

A Star Is Born! – p. 3/3

Rotational Quantities

A Star Is Born! – p. 4/3

Linear → Rotational Quantities

Linear Rotational Quantity Connection Quantity

s s = rθ θ = s

r

vT vT = rω ω = vT

r = dθ dt

a aT = rα α = aT

r = dω dt

KE = 1

2mv2

KER = 1

2Iω2

• F = m

a τ = rF⊥

• τ = I

α

• p = m

v

• L =

r × p

• L = I

ω

A Star Is Born! – p. 5/3

How Fast Will the Star Spin?

The pulsar in the Crab nebula has a period T0 = 0.033 s and this period has been observed to be increasing by ∆T = 1.26 × 10−5 s each year. Assuming constant angular acceleration what is the expression for the angular displacement of the pulsar? What are the values of the parameters in that expression? What is the torque exerted on the pulsar? mC = 3.4 × 1030 kg rC = 25 × 103 m

A Star Is Born! – p. 6/3

• F ∝

a → F = m a

Force and Motion 2

’Good’ data

A Star Is Born! – p. 7/3

Linear → Rotational Quantities

Linear Rotational Quantity Connection Quantity

s s = rθ θ = s

r

vT vT = rω ω = vT

r = dθ dt

a aT = rα α = aT

r = dω dt

KE = 1

2mv2

KER = 1

2Iω2

• F = m

a τ = rF⊥

• τ = I

α

• p = m

v

• L =

r × p

• L = I

ω

A Star Is Born! – p. 8/3

A Pulsar

A Star Is Born! – p. 9/3

Which One Wins?

A wooden disk and a metal ring have the same mass m and radius r and start from rest and roll down an inclined plane (see figure). What are the kinetic energies at the bottom in terms

• f the height of the incline h, m, r, and any other constants? Which one is going faster at the

bottom of the incline and gets to the bottom in the shortest time?

v v

d r

h

A Star Is Born! – p. 10/3

Moments of Inertia

A Star Is Born! – p. 11/3

Rolling Down an Incline - 1

A Star Is Born! – p. 12/3

Rolling Down an Incline - 2

40 20 20 40 40 20 20 40 x pixels y pixels Analysis of Rolling Down an Incline

A Star Is Born! – p. 13/3

Rolling Down an Incline - 3

ri

cm

r v

d

h

A Star Is Born! – p. 14/3

Which One Wins?

A wooden disk and a metal ring have the same mass m and radius r and start from rest and roll down an inclined plane (see figure). What are the kinetic energies at the bottom in terms

• f the height of the incline h, m, r, and any other constants? Which one is going faster at the

bottom of the incline and gets to the bottom in the shortest time?

v v

d r

h

A Star Is Born! – p. 15/3

Rolling Down an Incline

ri

cm

r v

d

h

A Star Is Born! – p. 16/3

Moments of Inertia

A Star Is Born! – p. 17/3

Rolling

A Star Is Born! – p. 18/3

Rotational Energy

Where is most of the Earth’s kinetic energy? Is it in the orbital motion around the Sun or in the spin about the Earth’s axis? Earth’s radius 6.37 × 106 m Earth-Sun distance 1.5 × 1011 m

A Star Is Born! – p. 19/3

Torque - Rotational Force

The shield door at a neutron test facil- ity at Lawrence Livermore Laboratory is possibly the world’s heaviest hinged

• door. It has a mass m = 44, 000 kg,

a rotational inertia about a vertical axis through its hinges of I = 8.7×104 kg − m2, and a (front) face width of w = 2.4 m. A steady force Fa = 73 N, ap- plied at its outer edge and perpendicu- lar to the plane of the door, can move it from rest through an angle θ = 90◦ in ∆t = 30 s. What is the torque ex- erted by the friction in the hinges? If the hinges have a radius rh = 0.1 m what is the friction force?

A Star Is Born! – p. 20/3

Torque - Rotational Equivalent of Force

Airplane String

A Star Is Born! – p. 21/3

Torque - Rotational Equivalent of Force

Airplane String

F

a

ra θ ra

a Airplane Propeller Radius Pivot R θ h Pivot Side View Top View

A Star Is Born! – p. 21/3

Torque - Rotational Equivalent of Force

Airplane String

F

a

ra θ ra

a Airplane Propeller Radius Pivot R θ h Pivot Side View Top View

F

a

ra Ff θ ra

a Airplane Propeller Radius Pivot R θ h Pivot Side View Top View

A Star Is Born! – p. 21/3

Torque - Rotational Equivalent of Force

• F = m

a → τ = r F⊥

F

A Star Is Born! – p. 22/3

Linear → Rotational Quantities

Linear Rotational Quantity Connection Quantity

s s = rθ θ = s

r

vT vT = rω ω = vT

r = dθ dt

aT aT = rα α = aT

r = dω dt

KE = 1

2mv2

KER = 1

2Iω2

• F = m

a τ = rF⊥

• τ = I

α

• p = m

v

• L =

r × p

• L = I

ω

A Star Is Born! – p. 23/3

Rotational Form of F = m a

Applied torque Disk Rotator

A Star Is Born! – p. 24/3

Torque and Rotational Energy - An Application

A trebuchet is a device used in the Middle Ages to throw big rocks at castles and is now used to throw other things like pumpkins, pianos, .... Consider the figures below. The trebuchet has a stiff wooden beam of mass mb = 15 kg and length lb = 5 m with masses mc = 700 kg (the counterweight) and mp = 0.1 kg (the payload) on it’s ends. Treat these two masses as point particles. A frictionless axle is located a distance d = 0.15 m from the

• counterweight. The beam is released from rest in a horizontal position. We will launch the

payload from a bucket at the end of the beam . What is the maximum speed the payload can reach before it leaves the bucket?

→

A Star Is Born! – p. 25/3

Collapsing Stars

Most stars in our galaxy will eventually run out of nuclear fuel and collapse to form a white dwarf star. The upper figure shows a white dwarf (the swall white dot at lower left) orbiting the star Sirius. The lower one shows a simu- lation of the explosion a white dwarf performed at the the University of Chicago. The yellow and orange represent the flame that pops out of the star, while the blue marks the surface of the star. The star is approximately the size

• f the Earth, but contains a mass greater than the Sun’s.

Suppose the Sun runs out of nuclear fuel and collapses into a white dwarf star with a radius equal to the radius

• f the Earth. What would be the rotation period Tf after

the collapse? Treat the stars as uniform spheres. Sun radius: 6.96 × 108 m Earth radius: 6.37 × 106 m TSun: 24.5 d

A Star Is Born! – p. 26/3

Angular Momentum

p p

||

p r x y φ

A Star Is Born! – p. 27/3

Twirling Student

A student volunteer/victim is spinning around on a turntable with her arms out- stretched. She is spinning initially at a rate of 0.6 revolutions/s and then drops her arms flat to her side at a distance rb = 0.20 m from the axis. What is her final rotation rate? Treat the student’s body as a cylinder with thin rods for arms. The turntable has a moment of inertia of It = 1 kg − m2. Arm length: 0.45 m Arm mass: 8 kg Cylinder mass: 55 kg

Arms Rotation Axis Body Turntable

A Star Is Born! – p. 28/3

Moments of Inertia

A Star Is Born! – p. 29/3

Linear → Rotational Quantities

Linear Rotational Quantity Connection Quantity

s s = rθ θ = s

r

vT v = rω ω = v

r = dθ dt

a a = rα α = a

r = dω dt

• F = m

a = d

p dt

• τ =

r × F

• τ = I

α = d

L dt

KE = 1

2mv2

KER = 1

2Iω2

• p = m

v

• L =

r × p

• L = I

ω

A Star Is Born! – p. 30/3

Angular Momentum Conservation

A Star Is Born! – p. 31/3

The Shape It’s In

The plot below shows the ‘obscuration’ in the angular area around B68 based on measurements of background stars. The light in the center is 1014 dimmer than outside the edge of the cloud. To make life simple we will treat the mass distribution of B68 as three, rigid, uniform spheres that lie along the axis shown in the figure and rotate with ω = 9.4 × 10−14 rad/s. The spheres do NOT rotate independently of the rest of the cloud. The origin is at the center of the central lobe. What is the moment of inertia of the cloud? Lobe Radius (km) Mass (kg) central Rc = 1.0 × 1012 mc = 6.0 × 1030 inner Ri = 2.0 × 1011 mi = 4.6 × 1028

• uter

Ro = 1.7 × 1011 mo = 2.9 × 1028

• rigin - inner cloud

center li = 1.4 × 1012 km

• rigin - outer cloud

center lo = 2.0 × 1012 km

• li

l

A Star Is Born! – p. 32/3

A Star Is Born

The molecular cloud B68 in the constellation Ophiuchus is rotating with an angular speed ω = 9.4 × 10−14 rad/s. The gravitational attraction among the atoms in the cloud may make it collapse until the core is hot enough to ignite nuclear reactions and B68 will begin to shine. If the final properties of B68 are the same as our Sun, i.e., the same mass and size, then what will be its final angular velocity and period? Assume the lost mass carries away very little angular momentum. Compare this with the angular velocity of the Sun. Is your result reasonable? Why or why not? MB68 = 6.04 × 1030 kg IB68 = 2.7 × 1054 kg − km2 MSun = 1.989 × 1030 kg RSun = 6.96 × 105 km TSun = 25.4 d

A Star Is Born! – p. 33/3

A Star Is Born

The molecular cloud B68 in the constellation Ophiuchus is rotating with an angular speed ω = 9.4 × 10−14 rad/s. The gravitational attraction among the atoms in the cloud may make it collapse until the core is hot enough to ignite nuclear reactions and B68 will begin to shine. If the final properties of B68 are the same as our Sun, i.e., the same mass and size, then what will be its final angular velocity and period? Assume the lost mass carries away very little angular momentum. Compare this with the angular velocity of the Sun. Is your result reasonable? Why or why not? MB68 = 6.04 × 1030 kg IB68 = 2.7 × 1054 kg − km2 MSun = 1.989 × 1030 kg RSun = 6.96 × 105 km TSun = 25.4 d

A Star Is Born! – p. 34/3

More Stars Are Born - The Eagle Nebula

The dark pillar-like structures are columns of cool interstellar hydrogen gas and dust that protrude from the interior wall of a dark molecular cloud. They have survived longer than their surround- ings in the face of a flood of ultraviolet light from hot, massive newborn stars (off the top edge of the picture). The tallest pillar (left) is about about 4 light-years long from base to tip. As the pillars are eroded away by the ultraviolet light, small globules of even denser gas buried within the pillars are uncovered. Forming inside at least some of the globules are embryonic stars. The picture was taken

• n April 1, 1995 with the Hubble Space

Telescope Wide Field and Planetary Camera 2. Credit: Jeff Hester and Paul Scowen (Ari- zona State University), and NASA/ESA.

A Star Is Born! – p. 35/3